Tutorial 6

Objectives Determine the solubility and solubility product of different salts. Compare the solubility of different salts. Define precipitation titrations. Define how to detect end point of argentometric titrations. Applications.

Solubility Solubility is the amount of substance that dissolves in a given volume of solvent at a given temperature, the solubility is expressed in mol/L. Solubility product constant, Ksp, is the constant for equilibrium expression representing the dissolving of an ionic solid in water. For a salt AaBb, the solubility product can be expressed as, Ksp = [A]a[B]b The [AaBb] was not considered in the expression as it is a pure solid.

Question 1: 1- The Ksp value for CaF2 is 4 x 10-11 at 25 degrees. Calculate its solubility at 25 degrees. CaF2 Ca+ + 2 F- S S 2 S Ksp = [aA]a[bB]b Ksp = [S]. [2S]2 = 4[S]3 S = 2.2 x 10-4 mol/L

Relative Solubilities: A salt’s Ksp value gives us information about its solubility, so IF: The salts being compared produce the same number of ions: We can compare the solubilities by comparing the Ksp values Directly. 2. The salts being compared produce different numbers of ions. The Ksp cannot be compared directly, So first we have to calculate the solubilities.

Question 2: Ksp Ag2CrO4=1.2X10-12 Ksp = [2S]2. [S]= 4S3 2- Justify that the ksp of Ag2CrO4 =1.2X10-12 is lower than AgCl=1.8x10-10, but Ag2CrO4 solubility is higher than AgCl . Ksp Ag2CrO4=1.2X10-12 Ag2CrO4 2Ag+ + CrO42- S 2S S Ksp = [2S]2. [S]= 4S3 S= 3 √( Ksp/4) = 6.69X10-5 mol/l  b) Ksp AgCl = 1.8x10-10 AgCl Ag+ + Cl- S S S Ksp = [S]. [S]= S2 S= √ Ksp = 1.34X10-5 mol/l

Precipitation titrations Precipitation titrations, is based on reactions that yields ionic compounds of limited solubility, precipitates. For a precipitate to be formed, the ionic product of its ions should be higher than the solubility product. If Ip < Ksp  Not ppt is formed If Ip > Ksp  ppt is formed

To calculate Ip we have to calculate the conc of Ag+ & Cl- Question 3: 3- 50 ml of 0.1 M NaCl is mixed with 50 ml of 0.2 M AgNO3. Predict whether a ppt of AgCl will be formed or not (Ksp of AgCl=1.8x10-10) To calculate Ip we have to calculate the conc of Ag+ & Cl- 1st conc of Cl- NaCl  Na+ + Cl- NaCl= 50 x 0.1= 5 mmoles [Cl-]= 5/100= 0.05 M

Ionic product = [Ag+][Cl-]= 0.1x0.05= 0.005 Question 3 2nd conc of Ag+ AgNO3 Ag+ + NO3- AgNO3=50 x 0.2= 10 mmoles [Ag+]=10/ 100 = 0.1 M Ionic product = [Ag+][Cl-]= 0.1x0.05= 0.005 Ip > Ksp AgCl 0.005 > 1.8x10-10 ppt is formed

Question 4 What minimum volume of 0.09621 M AgNO3 will be needed to assure an excess of silver ion in the titration of: A) NaCl sample that weighs 0.2513g (nMV)AgNO3 = (n Wt/Mwt) NaCl 1 x 0.09621 M x V = ( 1x 0.2513g ) /(23+35.5) V = 0.446 L = 44.64 ml

(nMV)AgNO3 = (n Wt/Mwt) Cl- Question 5 A Fajan titration of a 0.7908 g sample required 45.32 ml of 0.1046 M AgNO3 .Express the results of this analysis in term of the percentage of A) Cl- (nMV)AgNO3 = (n Wt/Mwt) Cl- 1x 0.1046 M x (45.32 x 10-3) = Wt/ 35.5 Wt = 0.168 g Wt%= 0.168 g / 0.7908 g x 100 = 21.28 %

Extra problems 1) What minimum volume of 0.09621 M AgNO3 will be needed to assure an excess of silver ion in the titration of: A) A 0.3462-g sample that is 74.52% (w/w) ZnCl2 B) 25.00 ml of 0.01907 M AlCl3 2) A Fajan titration of a 0.7908-g sample required 45.32 ml of 0.1046 M AgNO3 .Express the results of this analysis in term of the percentage of: A) BaCl2.2H2O B) ZnCl2.2NH4Cl(243.28 g/mol)